Mixing
Solution set of this linear system in reduced row echelon form.
Clash Royale CLAN TAG#URR8PPP
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So we have this matrix:
$beginbmatrix
1 & 0 & 2 & 1\
0 & 1 & -1 & 2
endbmatrix$
Does this matrix have infinitely many solution sets? How do I know? Is it because there are more variables than equations?
Here's my solution. Is this right?
let $ z = t$
so $y = 2 + t$
$x = 1 - 2t$
Is this right?
linear-algebra
edited Jul 31 at 20:20
Chickenmancer
2,937621
asked Jul 31 at 20:06
Jwan622
1,60111224
add a comment |Â
up vote
2
down vote
favorite
So we have this matrix:
$beginbmatrix
1 & 0 & 2 & 1\
0 & 1 & -1 & 2
endbmatrix$
Does this matrix have infinitely many solution sets? How do I know? Is it because there are more variables than equations?
Here's my solution. Is this right?
let $ z = t$
so $y = 2 + t$
$x = 1 - 2t$
Is this right?
linear-algebra
edited Jul 31 at 20:20
Chickenmancer
2,937621
asked Jul 31 at 20:06
Jwan622
1,60111224
Well, you have $$y = 2 + z \ x = 1 - 2 z$$ $z$ is a free variable, so you can call it $t$ and your result is correct.
– Moo
Jul 31 at 20:09
Yes, you're right.
– saulspatz
Jul 31 at 20:10
In the future, please take the time to enter important parts of your question as text instead of pasting pictures of them. Images are neither searchable nor accessible to screen readers. I find it hard to imagine that it would’ve taken you longer to type in that matrix and couple of lines of text than it did to take the picture, crop it, upload it and link it into your question.
– amd
Aug 1 at 0:13
A matrix doesn’t have a solution set; the system of equations that it might represent does.
– amd
Aug 1 at 0:13
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
So we have this matrix:
$beginbmatrix
1 & 0 & 2 & 1\
0 & 1 & -1 & 2
endbmatrix$
Does this matrix have infinitely many solution sets? How do I know? Is it because there are more variables than equations?
Here's my solution. Is this right?
let $ z = t$
so $y = 2 + t$
$x = 1 - 2t$
Is this right?
linear-algebra
edited Jul 31 at 20:20
Chickenmancer
2,937621
asked Jul 31 at 20:06
Jwan622
1,60111224
So we have this matrix:
$beginbmatrix
1 & 0 & 2 & 1\
0 & 1 & -1 & 2
endbmatrix$
Does this matrix have infinitely many solution sets? How do I know? Is it because there are more variables than equations?
Here's my solution. Is this right?
let $ z = t$
so $y = 2 + t$
$x = 1 - 2t$
Is this right?
linear-algebra
edited Jul 31 at 20:20
Chickenmancer
2,937621
asked Jul 31 at 20:06
Jwan622
1,60111224
edited Jul 31 at 20:20
Chickenmancer
2,937621
edited Jul 31 at 20:20
Chickenmancer
2,937621
edited Jul 31 at 20:20
Chickenmancer
2,937621
2,937621
asked Jul 31 at 20:06
Jwan622
1,60111224
asked Jul 31 at 20:06
Jwan622
1,60111224
asked Jul 31 at 20:06
Jwan622
1,60111224
1,60111224
Well, you have $$y = 2 + z \ x = 1 - 2 z$$ $z$ is a free variable, so you can call it $t$ and your result is correct.
– Moo
Jul 31 at 20:09
Yes, you're right.
– saulspatz
Jul 31 at 20:10
In the future, please take the time to enter important parts of your question as text instead of pasting pictures of them. Images are neither searchable nor accessible to screen readers. I find it hard to imagine that it would’ve taken you longer to type in that matrix and couple of lines of text than it did to take the picture, crop it, upload it and link it into your question.
– amd
Aug 1 at 0:13
A matrix doesn’t have a solution set; the system of equations that it might represent does.
– amd
Aug 1 at 0:13
add a comment |Â
Well, you have $$y = 2 + z \ x = 1 - 2 z$$ $z$ is a free variable, so you can call it $t$ and your result is correct.
– Moo
Jul 31 at 20:09
Yes, you're right.
– saulspatz
Jul 31 at 20:10
In the future, please take the time to enter important parts of your question as text instead of pasting pictures of them. Images are neither searchable nor accessible to screen readers. I find it hard to imagine that it would’ve taken you longer to type in that matrix and couple of lines of text than it did to take the picture, crop it, upload it and link it into your question.
– amd
Aug 1 at 0:13
A matrix doesn’t have a solution set; the system of equations that it might represent does.
– amd
Aug 1 at 0:13
Well, you have $$y = 2 + z \ x = 1 - 2 z$$ $z$ is a free variable, so you can call it $t$ and your result is correct.
– Moo
Jul 31 at 20:09
Well, you have $$y = 2 + z \ x = 1 - 2 z$$ $z$ is a free variable, so you can call it $t$ and your result is correct.
– Moo
Jul 31 at 20:09
Yes, you're right.
– saulspatz
Jul 31 at 20:10
Yes, you're right.
– saulspatz
Jul 31 at 20:10
In the future, please take the time to enter important parts of your question as text instead of pasting pictures of them. Images are neither searchable nor accessible to screen readers. I find it hard to imagine that it would’ve taken you longer to type in that matrix and couple of lines of text than it did to take the picture, crop it, upload it and link it into your question.
– amd
Aug 1 at 0:13
In the future, please take the time to enter important parts of your question as text instead of pasting pictures of them. Images are neither searchable nor accessible to screen readers. I find it hard to imagine that it would’ve taken you longer to type in that matrix and couple of lines of text than it did to take the picture, crop it, upload it and link it into your question.
– amd
Aug 1 at 0:13
A matrix doesn’t have a solution set; the system of equations that it might represent does.
– amd
Aug 1 at 0:13
A matrix doesn’t have a solution set; the system of equations that it might represent does.
– amd
Aug 1 at 0:13
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
We want to find all solutions to $x+2z=1$ and $y-z=2$. We can re-write this as $x=1-2z$ and $y=2+z$. Now it is clear that we can pick and number for $z$ and it will give us unique numbers for $x$ and $y$. Thus there are infinitely many solutions of the form
$$beginbmatrixx\y\z endbmatrix=beginbmatrix1-2t\2+t\t endbmatrix $$
for $tinBbb R$. So yes, your answer is correct.
In general an $mtimes n$ matrix is a map $Bbb R^nto Bbb R^m$. The rank tells you the dimension of the image, and the number of columns minus the rank tells you the dimension of the kernel. The dimension of the kernel is exactly the dimension of the number of solutions to a particular equation $Mx=b$, so in this case the solutions are $1$-dimensional, so there are infinitely many.
answered Jul 31 at 20:15
Elliot G
9,66521645
Is a system of 3 variables and two equations always with infinite solutions?
– Jwan622
Jul 31 at 20:34
"finitely many solutions" pretty sure it's a typo and you meant infinitely as your answer makes sense using the latter.
– WaveX
Jul 31 at 20:43
add a comment |Â
up vote
0
down vote
Let's solve the augmented matrix
$$
A=
beginbmatrix
1 & 0 & 2 & 1\
0 & 1 & -1 & 2
endbmatrix$$
i.e.
$$beginbmatrix
1 & 0 & 2 & 1\
0 & 1 & -1 & 2
endbmatrix
beginbmatrix
x \
y \
z \
-1
endbmatrix =
0$$
That is
$$begincases
x + 2z -1 &= 0 \
y -z -2 &= 0
endcases$$
Since this is a system of two equations in three unknowns, we have to parametrize in terms of one unknown, let's say $z = t$, then we get from the first equation
$$x = 1 -2t$$
and from the second equation
$$y = t+2$$
So the solution set is
$$lbrace
t in mathbbR:
beginbmatrix
1 -2t \
t+2\
t
endbmatrix
rbrace$$
Hence, your answer is correct.
answered Jul 31 at 20:25
Ahmad Bazzi
2,170417
Is a system of 3 variables and two equations always with infinite solutions?
– Jwan622
Jul 31 at 20:34
Yes since the rank of the matrix is at most $2$ (the original matrix or the augmented matrix by omitting the last column).
– Ahmad Bazzi
Jul 31 at 20:36
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
We want to find all solutions to $x+2z=1$ and $y-z=2$. We can re-write this as $x=1-2z$ and $y=2+z$. Now it is clear that we can pick and number for $z$ and it will give us unique numbers for $x$ and $y$. Thus there are infinitely many solutions of the form
$$beginbmatrixx\y\z endbmatrix=beginbmatrix1-2t\2+t\t endbmatrix $$
for $tinBbb R$. So yes, your answer is correct.
In general an $mtimes n$ matrix is a map $Bbb R^nto Bbb R^m$. The rank tells you the dimension of the image, and the number of columns minus the rank tells you the dimension of the kernel. The dimension of the kernel is exactly the dimension of the number of solutions to a particular equation $Mx=b$, so in this case the solutions are $1$-dimensional, so there are infinitely many.
answered Jul 31 at 20:15
Elliot G
9,66521645
Is a system of 3 variables and two equations always with infinite solutions?
– Jwan622
Jul 31 at 20:34
"finitely many solutions" pretty sure it's a typo and you meant infinitely as your answer makes sense using the latter.
– WaveX
Jul 31 at 20:43
add a comment |Â
up vote
2
down vote
We want to find all solutions to $x+2z=1$ and $y-z=2$. We can re-write this as $x=1-2z$ and $y=2+z$. Now it is clear that we can pick and number for $z$ and it will give us unique numbers for $x$ and $y$. Thus there are infinitely many solutions of the form
$$beginbmatrixx\y\z endbmatrix=beginbmatrix1-2t\2+t\t endbmatrix $$
for $tinBbb R$. So yes, your answer is correct.
In general an $mtimes n$ matrix is a map $Bbb R^nto Bbb R^m$. The rank tells you the dimension of the image, and the number of columns minus the rank tells you the dimension of the kernel. The dimension of the kernel is exactly the dimension of the number of solutions to a particular equation $Mx=b$, so in this case the solutions are $1$-dimensional, so there are infinitely many.
answered Jul 31 at 20:15
Elliot G
9,66521645
Is a system of 3 variables and two equations always with infinite solutions?
– Jwan622
Jul 31 at 20:34
"finitely many solutions" pretty sure it's a typo and you meant infinitely as your answer makes sense using the latter.
– WaveX
Jul 31 at 20:43
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We want to find all solutions to $x+2z=1$ and $y-z=2$. We can re-write this as $x=1-2z$ and $y=2+z$. Now it is clear that we can pick and number for $z$ and it will give us unique numbers for $x$ and $y$. Thus there are infinitely many solutions of the form
$$beginbmatrixx\y\z endbmatrix=beginbmatrix1-2t\2+t\t endbmatrix $$
for $tinBbb R$. So yes, your answer is correct.
In general an $mtimes n$ matrix is a map $Bbb R^nto Bbb R^m$. The rank tells you the dimension of the image, and the number of columns minus the rank tells you the dimension of the kernel. The dimension of the kernel is exactly the dimension of the number of solutions to a particular equation $Mx=b$, so in this case the solutions are $1$-dimensional, so there are infinitely many.
answered Jul 31 at 20:15
Elliot G
9,66521645
We want to find all solutions to $x+2z=1$ and $y-z=2$. We can re-write this as $x=1-2z$ and $y=2+z$. Now it is clear that we can pick and number for $z$ and it will give us unique numbers for $x$ and $y$. Thus there are infinitely many solutions of the form
$$beginbmatrixx\y\z endbmatrix=beginbmatrix1-2t\2+t\t endbmatrix $$
for $tinBbb R$. So yes, your answer is correct.
In general an $mtimes n$ matrix is a map $Bbb R^nto Bbb R^m$. The rank tells you the dimension of the image, and the number of columns minus the rank tells you the dimension of the kernel. The dimension of the kernel is exactly the dimension of the number of solutions to a particular equation $Mx=b$, so in this case the solutions are $1$-dimensional, so there are infinitely many.
answered Jul 31 at 20:15
Elliot G
9,66521645
edited Aug 1 at 4:04
answered Jul 31 at 20:15
Elliot G
9,66521645
answered Jul 31 at 20:15
Elliot G
9,66521645
answered Jul 31 at 20:15
Elliot G
9,66521645
9,66521645
Is a system of 3 variables and two equations always with infinite solutions?
– Jwan622
Jul 31 at 20:34
"finitely many solutions" pretty sure it's a typo and you meant infinitely as your answer makes sense using the latter.
– WaveX
Jul 31 at 20:43
add a comment |Â
Is a system of 3 variables and two equations always with infinite solutions?
– Jwan622
Jul 31 at 20:34
"finitely many solutions" pretty sure it's a typo and you meant infinitely as your answer makes sense using the latter.
– WaveX
Jul 31 at 20:43
Is a system of 3 variables and two equations always with infinite solutions?
– Jwan622
Jul 31 at 20:34
Is a system of 3 variables and two equations always with infinite solutions?
– Jwan622
Jul 31 at 20:34
"finitely many solutions" pretty sure it's a typo and you meant infinitely as your answer makes sense using the latter.
– WaveX
Jul 31 at 20:43
"finitely many solutions" pretty sure it's a typo and you meant infinitely as your answer makes sense using the latter.
– WaveX
Jul 31 at 20:43
add a comment |Â
up vote
0
down vote
Let's solve the augmented matrix
$$
A=
beginbmatrix
1 & 0 & 2 & 1\
0 & 1 & -1 & 2
endbmatrix$$
i.e.
$$beginbmatrix
1 & 0 & 2 & 1\
0 & 1 & -1 & 2
endbmatrix
beginbmatrix
x \
y \
z \
-1
endbmatrix =
0$$
That is
$$begincases
x + 2z -1 &= 0 \
y -z -2 &= 0
endcases$$
Since this is a system of two equations in three unknowns, we have to parametrize in terms of one unknown, let's say $z = t$, then we get from the first equation
$$x = 1 -2t$$
and from the second equation
$$y = t+2$$
So the solution set is
$$lbrace
t in mathbbR:
beginbmatrix
1 -2t \
t+2\
t
endbmatrix
rbrace$$
Hence, your answer is correct.
answered Jul 31 at 20:25
Ahmad Bazzi
2,170417
Is a system of 3 variables and two equations always with infinite solutions?
– Jwan622
Jul 31 at 20:34
Yes since the rank of the matrix is at most $2$ (the original matrix or the augmented matrix by omitting the last column).
– Ahmad Bazzi
Jul 31 at 20:36
add a comment |Â
up vote
0
down vote
Let's solve the augmented matrix
$$
A=
beginbmatrix
1 & 0 & 2 & 1\
0 & 1 & -1 & 2
endbmatrix$$
i.e.
$$beginbmatrix
1 & 0 & 2 & 1\
0 & 1 & -1 & 2
endbmatrix
beginbmatrix
x \
y \
z \
-1
endbmatrix =
0$$
That is
$$begincases
x + 2z -1 &= 0 \
y -z -2 &= 0
endcases$$
Since this is a system of two equations in three unknowns, we have to parametrize in terms of one unknown, let's say $z = t$, then we get from the first equation
$$x = 1 -2t$$
and from the second equation
$$y = t+2$$
So the solution set is
$$lbrace
t in mathbbR:
beginbmatrix
1 -2t \
t+2\
t
endbmatrix
rbrace$$
Hence, your answer is correct.
answered Jul 31 at 20:25
Ahmad Bazzi
2,170417
Is a system of 3 variables and two equations always with infinite solutions?
– Jwan622
Jul 31 at 20:34
Yes since the rank of the matrix is at most $2$ (the original matrix or the augmented matrix by omitting the last column).
– Ahmad Bazzi
Jul 31 at 20:36
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let's solve the augmented matrix
$$
A=
beginbmatrix
1 & 0 & 2 & 1\
0 & 1 & -1 & 2
endbmatrix$$
i.e.
$$beginbmatrix
1 & 0 & 2 & 1\
0 & 1 & -1 & 2
endbmatrix
beginbmatrix
x \
y \
z \
-1
endbmatrix =
0$$
That is
$$begincases
x + 2z -1 &= 0 \
y -z -2 &= 0
endcases$$
Since this is a system of two equations in three unknowns, we have to parametrize in terms of one unknown, let's say $z = t$, then we get from the first equation
$$x = 1 -2t$$
and from the second equation
$$y = t+2$$
So the solution set is
$$lbrace
t in mathbbR:
beginbmatrix
1 -2t \
t+2\
t
endbmatrix
rbrace$$
Hence, your answer is correct.
answered Jul 31 at 20:25
Ahmad Bazzi
2,170417
Let's solve the augmented matrix
$$
A=
beginbmatrix
1 & 0 & 2 & 1\
0 & 1 & -1 & 2
endbmatrix$$
i.e.
$$beginbmatrix
1 & 0 & 2 & 1\
0 & 1 & -1 & 2
endbmatrix
beginbmatrix
x \
y \
z \
-1
endbmatrix =
0$$
That is
$$begincases
x + 2z -1 &= 0 \
y -z -2 &= 0
endcases$$
Since this is a system of two equations in three unknowns, we have to parametrize in terms of one unknown, let's say $z = t$, then we get from the first equation
$$x = 1 -2t$$
and from the second equation
$$y = t+2$$
So the solution set is
$$lbrace
t in mathbbR:
beginbmatrix
1 -2t \
t+2\
t
endbmatrix
rbrace$$
Hence, your answer is correct.
answered Jul 31 at 20:25
Ahmad Bazzi
2,170417
answered Jul 31 at 20:25
Ahmad Bazzi
2,170417
answered Jul 31 at 20:25
Ahmad Bazzi
2,170417
answered Jul 31 at 20:25
Ahmad Bazzi
2,170417
2,170417
Is a system of 3 variables and two equations always with infinite solutions?
– Jwan622
Jul 31 at 20:34
Yes since the rank of the matrix is at most $2$ (the original matrix or the augmented matrix by omitting the last column).
– Ahmad Bazzi
Jul 31 at 20:36
add a comment |Â
Is a system of 3 variables and two equations always with infinite solutions?
– Jwan622
Jul 31 at 20:34
Yes since the rank of the matrix is at most $2$ (the original matrix or the augmented matrix by omitting the last column).
– Ahmad Bazzi
Jul 31 at 20:36
Is a system of 3 variables and two equations always with infinite solutions?
– Jwan622
Jul 31 at 20:34
Is a system of 3 variables and two equations always with infinite solutions?
– Jwan622
Jul 31 at 20:34
Yes since the rank of the matrix is at most $2$ (the original matrix or the augmented matrix by omitting the last column).
– Ahmad Bazzi
Jul 31 at 20:36
Yes since the rank of the matrix is at most $2$ (the original matrix or the augmented matrix by omitting the last column).
– Ahmad Bazzi
Jul 31 at 20:36
add a comment |Â
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Well, you have $$y = 2 + z \ x = 1 - 2 z$$ $z$ is a free variable, so you can call it $t$ and your result is correct.
– Moo
Jul 31 at 20:09
Yes, you're right.
– saulspatz
Jul 31 at 20:10
In the future, please take the time to enter important parts of your question as text instead of pasting pictures of them. Images are neither searchable nor accessible to screen readers. I find it hard to imagine that it would’ve taken you longer to type in that matrix and couple of lines of text than it did to take the picture, crop it, upload it and link it into your question.
– amd
Aug 1 at 0:13
A matrix doesn’t have a solution set; the system of equations that it might represent does.
– amd
Aug 1 at 0:13